Q: Series involving log $\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$ Does anybody know how to prove this series?
$$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$
I arrived at this through Mathematica.
I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.

A: Note that
\begin{align*}
\sum_{n=1}^{\infty}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right)
&=\lim_{N\to\infty} \sum_{n=1}^{N}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right)\\
&=\lim_{N\to\infty} \log\left[ e^{-N} \prod_{n=1}^{N} \left(\frac{2n+1}{2n-1}\right)^{n} \right] \\
&=\lim_{N\to\infty} \log\left[ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \right].
\end{align*}
By Stirling's formula, it follows that
$$ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \sim \sqrt{\frac{e}{2}}. $$
This immediately yields the desired answer.

A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}\bracks{n\ \ln\pars{2n + 1 \over 2n - 1} - 1}
     ={1 - \ln\pars{2} \over 2}}$

With ${\ds{\quad N \in {\mathbb N}\,,\quad N \geq 1}}$:
  \begin{align}&\color{#c00000}{\sum_{n = 1}^{N}n\ \ln\pars{2n + 1 \over 2n - 1}}
=\sum_{n = 1}^{N}n\int_{-1}^{1}{\dd x \over x + 2n}
=\half\int_{0}^{1}\sum_{n = 1}^{N}\pars{{n \over n + x/2} + {n \over n - x/2}}
\,\dd x
\\[3mm]&=\half\int_{0}^{1}\sum_{n = 1}^{N}
\pars{1 - {x/2 \over n + x/2} + 1 + {x/2 \over n - x/2}}\,\dd x
\\[3mm]&=N + {1 \over 4}\int_{0}^{1}
x\sum_{n = 1}^{N}{x \over \pars{n + x/2}\pars{n - x/2}}\,\dd x
\\[3mm]&=N + {1 \over 4}\int_{0}^{1}
x\sum_{n = 0}^{N - 1}{x \over \pars{n + 1 + x/2}\pars{n + 1 - x/2}}\,\dd x
\end{align}

$$
\color{#c00000}{\sum_{n = 1}^{N}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}}
={1 \over 4}\int_{0}^{1}
x\sum_{n = 0}^{N - 1}{x \over \pars{n + 1 + x/2}\pars{n + 1 - x/2}}\,\dd x
$$

\begin{align}
&\color{#c00000}{\sum_{n = 1}^{\infty}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}}
={1 \over 4}\int_{0}^{1}x\bracks{%
\Psi\pars{1 + {x \over 2}} - \Psi\pars{1 - {x \over 2}}}\,\dd x
\\[3mm]&={1 \over 4}\int_{0}^{1}x\bracks{%
{2 \over x} + \Psi\pars{x \over 2} - \Psi\pars{1 - {x \over 2}}}\,\dd x
=\half - {1 \over 4}\int_{0}^{1}x\pi\cot\pars{\pi x \over 2}\,\dd x
\end{align}

$$
\color{#c00000}{\sum_{n = 1}^{\infty}\bracks{n\ln\pars{2n + 1 \over 2n - 1} - 1}}
=\half - {1 \over \pi}\color{#00f}{\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x}\tag{1}
$$

\begin{align}&\color{#00f}{\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x}
=\left.x\ln\pars{\sin\pars{x}}\vphantom{\LARGE A}\right\vert_{0}^{\pi/2}
-\
\underbrace{\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x}
_{\ds{-\,{\pi\ln\pars{2} \over 2}}}\ =\
\color{#00f}{{\pi\ln\pars{2} \over 2}}
\end{align}
  The $\ds{\large\ln\pars{\sin\pars{\cdots}}}$-integral is reported by M.SE as
  a frequent question. Replacing the last result in $\pars{1}$:

$$\color{#66f}{\large%
\sum_{n = 1}^{\infty}\bracks{n\ \ln\pars{2n + 1 \over 2n - 1} - 1}
={1 - \ln\pars{2} \over 2}} \approx 0.1534
$$

A: $$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$
\begin{align}
\log(1+x) - \log(1-x) & = \left(x - \dfrac{x^2}2 + \dfrac{x^3}3 \mp \cdots\right) - \left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots\right)\\
& = 2\left(x + \dfrac{x^3}3 + \dfrac{x^5}5 + \cdots \right)
\end{align}
Hence,
$$n \left(\log(1+1/2n) - \log(1-1/2n)\right) = 2n \left(\sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k+1}}\right) = \sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$
Hence,
$$n \log \left(\dfrac{2n+1}{2n-1}\right) - 1 = \sum_{k=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$
$$\sum_{n=1}^{\infty} \left(n \log \left(\dfrac{2n+1}{2n-1}\right) - 1\right) = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}} = \sum_{k=1}^{\infty} \dfrac{\zeta(2k)}{(2k+1) \cdot 2^{2k}}$$
\begin{align}f(m) & = \sum_{n=1}^m \left(n \log\left(\dfrac{2n+1}{2n-1}\right)-1\right)\\
& = \left( 1\log(3) - 1 + 2 \log(5) - 2 \log(3) + 1 + \cdots \right)\\
& = m - \log(3) - \log(5) - \cdots - \log(2m-1) + m \log(2m+1)
\end{align}
Now
$$\log(1) + \log(3) + \log(5) + \cdots + \log(2m-1) = \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$
Hence,
$$f(m) = m + m \log(2m+1) - \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$
Now use Stirling to get your answer.
